BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Point on the line

Expert replies
Source: — Data Sufficiency |

by Matt@VeritasPrep » Thu Aug 08, 2013 4:04 pm
The point (r,s) has an x-coordinate of r and a y-coordinate of s. It will be on the line y = 3x + 2 if its coordinates satisfy that equation; in other words, if s = 3r + 2.

S1::

(3r + 2 - s) * (4r + 9 - s) = 0

implies that either

(3r + 2 - s) = 0
or
(4r + 9 - s) = 0

If (3r + 2) - s = 0, then (3r + 2) = s ... on the line!
If (4r + 9) - s = 0, then (4r + 9) = s ... NOT (necessarily) on the line

We don't know which is true, so insufficient.

S2::

Same as S1 - two equations, one of which gives s = 3r + 2, the other of which gives s = 4r - 6. (r,s) could be on the line, or it could not be. Insufficient again.

S1 + S2::

Either (3r + 2 - s) = 0, or it does not, in which case BOTH (4r + 9 - s) and (4r - 6 - s) = 0.

But if (4r + 9 - s) = (4r - 6 - s) = 0, then 4r + 9 = s and 4r - 6 = s, meaning that 4r + 9 = 4r - 6. That equation is IMPOSSIBLE for any value of r. So (4r + 9 - s) and (4r - 6 - s) can't equal 0.

Thus (3r + 2 - s) must equal 0, and (r,s) is on the line!

Let me know if I can fill in any more details on this; the interpretation can be tricky.
Join the discussion

by GMATGuruNY » Thu Aug 08, 2013 5:08 pm

In the xy-plane, does the line with equation y = 3x + 2 contain the point (r, s)?

1) (3r+2-s)(4r+9-s) = 0
2) (4r-6-s)(3r+2-s) = 0
If (r,s) is a point on the line y = 3x + 2, then s = 3r + 2, implying that 3r - s = -2.

Question stem rephrased: Does 3r - s = -2?

Statement 1: (3r+2-s)(4r+9-s) = 0
Either 3r+2-s = 0 or 4r+9-s = 0.
If 3r+2-s = 0, then 3r - s = -2.
If 4r+9-s = 0, then 4r - s = -9, in which case it cannot be determined whether 3r - s = -2.
INSUFFICIENT.

Statement 2: (4r-6-s)(3r+2-s) = 0
Either 4r-6-s=0 or 3r+2-s = 0.
If 3r+2-s = 0, then 3r - s = -2.
If 4r-6-s = 0, then 4r - s = 6, in which case it cannot be determined whether 3r - s = -2.
INSUFFICIENT.

Statements 1 and 2 combined:
4r - s = -9 (from statement 1) and 4r - s = 6 (from statement 2) cannot both be true, since 4r - s cannot be equal to more than one value.
Thus, the only way the equations in the two statements can both be equal to 0 is if 3r - s = -2.
SUFFICIENT.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by Brent@GMATPrepNow » Thu Aug 08, 2013 9:35 pm
satish_iitg wrote:In the xy plane, does the line with equation y = 3x+2 contain the point (r,s)

1) (3r+2-s) (4r+9-s) = 0

1) (4r-6-s) (3r+2-s) = 0
If (r,s) is on the line defined by the equation y=3x+2, then (r,s) must satisfy the equation y=3x+2. In other words, it must be true that s=3r+2
For example: We know that the point (5, 17) is on the line y=3x+2, because when we plug x=5 and y=17 into the equation, we get 17 = 3(5)+2 and the equation holds true.

So, we can reword the target question to be "Does s = 3r + 2?"

1. (3r+2-s)(4r+9-s) = 0
From this, we know that either (3r+2-s) = 0 or (4r+9-s) = 0
If (3r+2-s) = 0 then s = 3r+2, in which case the answer to our new target question is yes
If (4r+9-s) = 0 then s = 4r+9, in which case the answer to our new target question is no
Since we get two different answers to the target question, statement 1 is NOT SUFFICIENT

2. (4r-6-s)(3r+2-s) = 0
From this, we know that either (4r-6-s) = 0 or (3r+2-s) = 0
If (4r-6-s)) = 0 then s = 4r-6, in which case the answer to our new target question is no
If (3r+2-s) = 0 then s = 3r+2, in which case the answer to our new target question is yes
Since we get two different answers to the target question, statement 2 is NOT SUFFICIENT

Statements 1&2 combined: Since (3r+2-s) is the only expression common to both statements, it must be true that 3r+2-s = 0, in which case s MUST equal 3r+2
As such the answer is C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by satish_iitg » Thu Aug 08, 2013 11:37 pm
Thanks every one. this was second question on my Gmat Prep test.
Join the discussion